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## Homework Statement

Hey all,

I am having trouble following some of the notes that my professor posted with regards to waves inside a blackbody; here is what he posted: (the part in bold is what I am just not understanding)

"Inside the blackbody box, we need for the position of the walls of the box to be a node for each dimension. For a one-dimensional cavity, e.g., we would choose x = 0 at one wall and then require the other end to be 2a/λ = n (where a is the length of of a side of the cube, which he defined as the shape of this particular blackbody).

**How many such waves can we fit inside a box of dimensions a^3?**We can figure out how to calculate this geometrically by looking at a cube of points in a 3-dimensional space where each point represents a coordinate deﬁned by (nx, ny, nz) and nx = 1, 2, 3, ... and the same for ny and nz. This is a graph in which only positive integer values for nx, ny, and nz are allowed and we want to know the size of the cube (i.e. the number of points deﬁned by the coordinates we just deﬁned) that ﬁt inside of the 1/8 of a sphere whose radius is r = (2a/λ) with shell thickness givenby dr = (2a/λ^2)dλ. The

**net result of the density of states calculation is**

N(λ)dλ= (2πa^3)/(cλ^2) dλ

N(λ)dλ= (2πa^3)/(cλ^2) dλ

I have absolutely no idea how he made the jump to that result, and he didn't indicate any other calculations in the note page.

## The Attempt at a Solution

Since there was a dλ in the density of states he arrived at, I thought he might have taken the dV/dr of a sphere such that dv=4πr^2 dr

As he noted in the paragraph above, we should only be looking in 1/8 of this region, so I multiplied this dv*1/8 (which is the same effect as just multiplying the original volume by 1/8...) and got

dv=(πr^2)/2 dr

I then substituted dr=(2a/λ^2) as indicated in the above paragraph and got:

dV=aπr

^{2}a/λ

^{2}

I then substituted in r

^{2}=4a

^{2}/λ

^{2}...

What I did is clearly wrong. I have no idea how he arrived at having a 2 in the numerator rather than a 4, or how he got a c in the denominator. Any help

*of you could give me would be HUGELY appreciated.*

**any**Thanks